A Simple Solution is to use two loops to find XOR of all subarrays and return the maximum.

C++

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// A simple C++ program to find max subarray XOR

#include<bits/stdc++.h>

usingnamespacestd;

intmaxSubarrayXOR(intarr[], intn)

{

intans = INT_MIN; // Initialize result

// Pick starting points of subarrays

for(inti=0; i<n; i++)

{

intcurr_xor = 0; // to store xor of current subarray

// Pick ending points of subarrays starting with i

for(intj=i; j<n; j++)

{

curr_xor = curr_xor ^ arr[j];

ans = max(ans, curr_xor);

}

}

returnans;

}

// Driver program to test above functions

intmain()

{

intarr[] = {8, 1, 2, 12};

intn = sizeof(arr)/sizeof(arr[0]);

cout << "Max subarray XOR is "<< maxSubarrayXOR(arr, n);

return0;

}

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Java

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// A simple Java program to find max subarray XOR

classGFG {

staticintmaxSubarrayXOR(intarr[], intn)

{

intans = Integer.MIN_VALUE; // Initialize result

// Pick starting points of subarrays

for(inti=0; i<n; i++)

{

// to store xor of current subarray

intcurr_xor = 0;

// Pick ending points of subarrays starting with i

for(intj=i; j<n; j++)

{

curr_xor = curr_xor ^ arr[j];

ans = Math.max(ans, curr_xor);

}

}

returnans;

}

// Driver program to test above functions

publicstaticvoidmain(String args[])

{

intarr[] = {8, 1, 2, 12};

intn = arr.length;

System.out.println("Max subarray XOR is "+

maxSubarrayXOR(arr, n));

}

}

//This code is contributed by Sumit Ghosh

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Python3

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# A simple Python program

# to find max subarray XOR

defmaxSubarrayXOR(arr,n):

ans =-2147483648#Initialize result

# Pick starting points of subarrays

fori inrange(n):

# to store xor of current subarray

curr_xor =0

# Pick ending points of

# subarrays starting with i

forj inrange(i,n):

curr_xor =curr_xor ^ arr[j]

ans =max(ans, curr_xor)

returnans

# Driver code

arr =[8, 1, 2, 12]

n =len(arr)

print("Max subarray XOR is ",

maxSubarrayXOR(arr, n))

# This code is contributed

# by Anant Agarwal.

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C#

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// A simple C# program to find

// max subarray XOR

usingSystem;

classGFG

{

// Function to find max subarray

staticintmaxSubarrayXOR(int[]arr, intn)

{

intans = int.MinValue;

// Initialize result

// Pick starting points of subarrays

for(inti = 0; i < n; i++)

{

// to store xor of current subarray

intcurr_xor = 0;

// Pick ending points of

// subarrays starting with i

for(intj = i; j < n; j++)

{

curr_xor = curr_xor ^ arr[j];

ans = Math.Max(ans, curr_xor);

}

}

returnans;

}

// Driver code

publicstaticvoidMain()

{

int[]arr = {8, 1, 2, 12};

intn = arr.Length;

Console.WriteLine("Max subarray XOR is "+

maxSubarrayXOR(arr, n));

}

}

// This code is contributed by Sam007.

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PHP

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<?php

// A simple PHP program to

// find max subarray XOR

functionmaxSubarrayXOR($arr, $n)

{

// Initialize result

$ans= PHP_INT_MIN;

// Pick starting points

// of subarrays

for($i= 0; $i< $n; $i++)

{

// to store xor of

// current subarray

$curr_xor= 0;

// Pick ending points of

// subarrays starting with i

for($j= $i; $j< $n; $j++)

{

$curr_xor= $curr_xor^ $arr[$j];

$ans= max($ans, $curr_xor);

}

}

return$ans;

}

// Driver Code

$arr= array(8, 1, 2, 12);

$n= count($arr);

echo"Max subarray XOR is "

, maxSubarrayXOR($arr, $n);

// This code is contributed by anuj_67.

?>

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Output:

Max subarray XOR is 15

Time Complexity of above solution is O(n2).

An Efficient Solution can solve the above problem in O(n) time under the assumption that integers take fixed number of bits to store. The idea is to use Trie Data Structure. Below is algorithm.

1) Create an empty Trie. Every node of Trie is going to
contain two children, for 0 and 1 value of bit.
2) Initialize pre_xor = 0 and insert into the Trie.
3) Initialize result = minus infinite
4) Traverse the given array and do following for every
array element arr[i].
a) pre_xor = pre_xor ^ arr[i]
pre_xor now contains xor of elements from
arr[0] to arr[i].
b) Query the maximum xor value ending with arr[i]
from Trie.
c) Update result if the value obtained in step
4.b is more than current value of result.

How does 4.b work?
We can observe from above algorithm that we build a Trie that contains XOR of all prefixes of given array. To find the maximum XOR subarray ending with arr[i], there may be two cases.
i) The prefix itself has the maximum XOR value ending with arr[i]. For example if i=2 in {8, 2, 1, 12}, then the maximum subarray xor ending with arr[2] is the whole prefix.
ii) We need to remove some prefix (ending at index from 0 to i-1). For example if i=3 in {8, 2, 1, 12}, then the maximum subarray xor ending with arr[3] starts with arr[1] and we need to remove arr[0].

To find the prefix to be removed, we find the entry in Trie that has maximum XOR value with current prefix. If we do XOR of such previous prefix with current prefix, we get the maximum XOR value ending with arr[i].
If there is no prefix to be removed (case i), then we return 0 (that’s why we inserted 0 in Trie).

Below is the implementation of above idea :

C++

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// C++ program for a Trie based O(n) solution to find max

// subarray XOR

#include<bits/stdc++.h>

usingnamespacestd;

// Assumed int size

#define INT_SIZE 32

// A Trie Node

structTrieNode

{

intvalue; // Only used in leaf nodes

TrieNode *arr[2];

};

// Utility function tp create a Trie node

TrieNode *newNode()

{

TrieNode *temp = newTrieNode;

temp->value = 0;

temp->arr[0] = temp->arr[1] = NULL;

returntemp;

}

// Inserts pre_xor to trie with given root

voidinsert(TrieNode *root, intpre_xor)

{

TrieNode *temp = root;

// Start from the msb, insert all bits of

// pre_xor into Trie

for(inti=INT_SIZE-1; i>=0; i--)

{

// Find current bit in given prefix

boolval = pre_xor & (1<<i);

// Create a new node if needed

if(temp->arr[val] == NULL)

temp->arr[val] = newNode();

temp = temp->arr[val];

}

// Store value at leaf node

temp->value = pre_xor;

}

// Finds the maximum XOR ending with last number in

// prefix XOR 'pre_xor' and returns the XOR of this maximum

// with pre_xor which is maximum XOR ending with last element

// of pre_xor.

intquery(TrieNode *root, intpre_xor)

{

TrieNode *temp = root;

for(inti=INT_SIZE-1; i>=0; i--)

{

// Find current bit in given prefix

boolval = pre_xor & (1<<i);

// Traverse Trie, first look for a

// prefix that has opposite bit

if(temp->arr[1-val]!=NULL)

temp = temp->arr[1-val];

// If there is no prefix with opposite

// bit, then look for same bit.

elseif(temp->arr[val] != NULL)

temp = temp->arr[val];

}

returnpre_xor^(temp->value);

}

// Returns maximum XOR value of a subarray in arr[0..n-1]

intmaxSubarrayXOR(intarr[], intn)

{

// Create a Trie and insert 0 into it

TrieNode *root = newNode();

insert(root, 0);

// Initialize answer and xor of current prefix

intresult = INT_MIN, pre_xor =0;

// Traverse all input array element

for(inti=0; i<n; i++)

{

// update current prefix xor and insert it into Trie

pre_xor = pre_xor^arr[i];

insert(root, pre_xor);

// Query for current prefix xor in Trie and update

// result if required

result = max(result, query(root, pre_xor));

}

returnresult;

}

// Driver program to test above functions

intmain()

{

intarr[] = {8, 1, 2, 12};

intn = sizeof(arr)/sizeof(arr[0]);

cout << "Max subarray XOR is "<< maxSubarrayXOR(arr, n);

return0;

}

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Java

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// Java program for a Trie based O(n) solution to

// find max subarray XOR

classGFG

{

// Assumed int size

staticfinalintINT_SIZE = 32;

// A Trie Node

staticclassTrieNode

{

intvalue; // Only used in leaf nodes

TrieNode[] arr = newTrieNode[2];

publicTrieNode() {

value = 0;

arr[0] = null;

arr[1] = null;

}

}

staticTrieNode root;

// Inserts pre_xor to trie with given root

staticvoidinsert(intpre_xor)

{

TrieNode temp = root;

// Start from the msb, insert all bits of

// pre_xor into Trie

for(inti=INT_SIZE-1; i>=0; i--)

{

// Find current bit in given prefix

intval = (pre_xor & (1<<i)) >=1? 1: 0;

// Create a new node if needed

if(temp.arr[val] == null)

temp.arr[val] = newTrieNode();

temp = temp.arr[val];

}

// Store value at leaf node

temp.value = pre_xor;

}

// Finds the maximum XOR ending with last number in

// prefix XOR 'pre_xor' and returns the XOR of this

// maximum with pre_xor which is maximum XOR ending

// with last element of pre_xor.

staticintquery(intpre_xor)

{

TrieNode temp = root;

for(inti=INT_SIZE-1; i>=0; i--)

{

// Find current bit in given prefix

intval = (pre_xor & (1<<i)) >= 1? 1: 0;

// Traverse Trie, first look for a

// prefix that has opposite bit

if(temp.arr[1-val] != null)

temp = temp.arr[1-val];

// If there is no prefix with opposite

// bit, then look for same bit.

elseif(temp.arr[val] != null)

temp = temp.arr[val];

}

returnpre_xor^(temp.value);

}

// Returns maximum XOR value of a subarray in

// arr[0..n-1]

staticintmaxSubarrayXOR(intarr[], intn)

{

// Create a Trie and insert 0 into it

root = newTrieNode();

insert(0);

// Initialize answer and xor of current prefix

intresult = Integer.MIN_VALUE;

intpre_xor =0;

// Traverse all input array element

for(inti=0; i<n; i++)

{

// update current prefix xor and insert it

// into Trie

pre_xor = pre_xor^arr[i];

insert(pre_xor);

// Query for current prefix xor in Trie and

// update result if required

result = Math.max(result, query(pre_xor));

}

returnresult;

}

// Driver program to test above functions

publicstaticvoidmain(String args[])

{

intarr[] = {8, 1, 2, 12};

intn = arr.length;

System.out.println("Max subarray XOR is "+

maxSubarrayXOR(arr, n));

}

}

// This code is contributed by Sumit Ghosh

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C#

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usingSystem;

// C# program for a Trie based O(n) solution to

// find max subarray XOR

publicclassGFG

{

// Assumed int size

publicconstintINT_SIZE = 32;

// A Trie Node

publicclassTrieNode

{

publicintvalue; // Only used in leaf nodes

publicTrieNode[] arr = newTrieNode[2];

publicTrieNode()

{

value = 0;

arr[0] = null;

arr[1] = null;

}

}

publicstaticTrieNode root;

// Inserts pre_xor to trie with given root

publicstaticvoidinsert(intpre_xor)

{

TrieNode temp = root;

// Start from the msb, insert all bits of

// pre_xor into Trie

for(inti = INT_SIZE-1; i >= 0; i--)

{

// Find current bit in given prefix

intval = (pre_xor & (1 << i)) >= 1 ? 1 : 0;

// Create a new node if needed

if(temp.arr[val] == null)

{

temp.arr[val] = newTrieNode();

}

temp = temp.arr[val];

}

// Store value at leaf node

temp.value = pre_xor;

}

// Finds the maximum XOR ending with last number in

// prefix XOR 'pre_xor' and returns the XOR of this

// maximum with pre_xor which is maximum XOR ending

// with last element of pre_xor.

publicstaticintquery(intpre_xor)

{

TrieNode temp = root;

for(inti = INT_SIZE-1; i >= 0; i--)

{

// Find current bit in given prefix

intval = (pre_xor & (1 << i)) >= 1 ? 1 : 0;

// Traverse Trie, first look for a

// prefix that has opposite bit

if(temp.arr[1 - val] != null)

{

temp = temp.arr[1 - val];

}

// If there is no prefix with opposite

// bit, then look for same bit.

elseif(temp.arr[val] != null)

{

temp = temp.arr[val];

}

}

returnpre_xor ^ (temp.value);

}

// Returns maximum XOR value of a subarray in

// arr[0..n-1]

publicstaticintmaxSubarrayXOR(int[] arr, intn)

{

// Create a Trie and insert 0 into it

root = newTrieNode();

insert(0);

// Initialize answer and xor of current prefix

intresult = int.MinValue;

intpre_xor = 0;

// Traverse all input array element

for(inti = 0; i < n; i++)

{

// update current prefix xor and insert it

// into Trie

pre_xor = pre_xor ^ arr[i];

insert(pre_xor);

// Query for current prefix xor in Trie and

// update result if required

result = Math.Max(result, query(pre_xor));

}

returnresult;

}

// Driver program to test above functions

publicstaticvoidMain(string[] args)

{

int[] arr = newint[] {8, 1, 2, 12};

intn = arr.Length;

Console.WriteLine("Max subarray XOR is "+ maxSubarrayXOR(arr, n));

}

}

// This code is contributed by Shrikant13

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Output:

Max subarray XOR is 15

Exercise: Extend the above solution so that it also prints starting and ending indexes of subarray with maximum value (Hint: we can add one more field to Trie node to achieve this)

This article is contributed by Romil Punetha. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.